Consider the parallel $LCR$ circuit shown in the figure. Find the net current $i$ and the phase of $i$. Show that $i = \frac{V}{Z}$. Find the impedance $Z$ for this circuit.

(N/A) The circuit consists of a resistor $R$ in parallel with a series combination of a capacitor $C$ and an inductor $L$. The voltage across both branches is $V = V_m \sin \omega t$.
$1$. Current through the resistor branch $(i_R)$:
$i_R = \frac{V}{R} = \frac{V_m}{R} \sin \omega t$.
$2$. Current through the $LC$ branch $(i_{LC})$:
The impedance of the $LC$ series branch is $Z_{LC} = j(\omega L - \frac{1}{\omega C})$.
The current is $i_{LC} = \frac{V}{Z_{LC}} = \frac{V_m \sin \omega t}{j(\omega L - \frac{1}{\omega C})} = \frac{V_m \sin(\omega t - \pi/2)}{\omega L - 1/(\omega C)}$ (if $\omega L > 1/\omega C$).
$3$. Total current $i$:
$i = i_R + i_{LC} = \frac{V_m}{R} \sin \omega t + \frac{V_m}{\omega L - 1/(\omega C)} \sin(\omega t - \pi/2)$.
Using phasor addition,the total current is $i = I_m \sin(\omega t + \phi)$,where $I_m = V_m \sqrt{(\frac{1}{R})^2 + (\frac{1}{\omega L - 1/(\omega C)})^2}$.
$4$. Impedance $Z$:
Since $i = V/Z$,the total impedance $Z$ is given by $\frac{1}{Z} = \sqrt{(\frac{1}{R})^2 + (\frac{1}{\omega L - 1/(\omega C)})^2}$.
Therefore,$Z = \frac{R |\omega L - 1/(\omega C)|}{\sqrt{R^2 + (\omega L - 1/(\omega C))^2}}$.